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3 years ago

Which expression represents the series

Mathematics

2 answers:

kompoz [17]3 years ago

6 0

That would be the first choice:

5^0 = 1 , 5^1 = 5, 5^2 = 25 and so on up to 5^4 = 625

Umnica [9.8K]3 years ago

6 0

Answer:

The first option is the correct answer.

Step-by-step explanation:

The given series is :

1 + 5 + 25 + 125 + 625

We get ;

$5^0=1$

$5^1=5$

$5^2=25$

$5^3=125$

$5^4=625$

We can see that the power/exponent is ranging from 0 to 4.

So, the first option is the right answer where 'i' goes from 0 to 4.

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Answer:

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Step-by-step explanation:

two negatives equals positive

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3 years ago

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(i) 3 (a + 5b) + a (a + 4) (ii) y (10 - y) + 3 (y - 2) (iii) 2 (8a - 5b) + 3 (5a - 12) (iv) 3 (y - 3) + ( 8 - 6y + x) (v) a (a -

AveGali [126]

Answer:

$3 (a + 5b) + a (a + 4) = a^2 + 7a + 15b$

$y (10 - y) + 3 (y - 2) = - y^2+13y - 6$

$2 (8a - 5b) + 3 (5a - 12) = 31a -10b - 36$

$3 (y - 3) + ( 8 - 6y + x) = -3y + x-1$

$a (a - 2b) + b (b + 2a - c) =a^2 + b^2 - bc$

$5 (x - y + z) + (4x + 3y) =9x - 2y +5z$

Step-by-step explanation:

Solving (i):

$3 (a + 5b) + a (a + 4)$

Open brackets

$3 (a + 5b) + a (a + 4) = 3a + 15b + a^2 + 4a$

Collect like terms

$3 (a + 5b) + a (a + 4) = a^2 + 4a+3a + 15b$

$3 (a + 5b) + a (a + 4) = a^2 + 7a + 15b$

Solving (ii)

$y (10 - y) + 3 (y - 2)$

Open bracket

$y (10 - y) + 3 (y - 2) = 10y - y^2 + 3y - 6$

Collect like terms

$y (10 - y) + 3 (y - 2) = - y^2+10y + 3y - 6$

$y (10 - y) + 3 (y - 2) = - y^2+13y - 6$

Solving (iii)

$2 (8a - 5b) + 3 (5a - 12)$

Open bracket

$2 (8a - 5b) + 3 (5a - 12) = 16a -10b + 15a - 36$

Collect like terms

$2 (8a - 5b) + 3 (5a - 12) = 16a+ 15a -10b - 36$

$2 (8a - 5b) + 3 (5a - 12) = 31a -10b - 36$

Solving (iv)

$3 (y - 3) + ( 8 - 6y + x)$

Open bracket

$3 (y - 3) + ( 8 - 6y + x) = 3y - 9 + 8 - 6y + x$

Collect like terms

$3 (y - 3) + ( 8 - 6y + x) = 3y - 6y + x- 9 + 8$

$3 (y - 3) + ( 8 - 6y + x) = -3y + x-1$

Solving (v):

$a (a - 2b) + b (b + 2a - c)$

Open bracket

$a (a - 2b) + b (b + 2a - c) =a^2 - 2ab + b^2 + 2ab - bc$

Collect like terms

$a (a - 2b) + b (b + 2a - c) =a^2 + 2ab- 2ab + b^2 - bc$

$a (a - 2b) + b (b + 2a - c) =a^2 + b^2 - bc$

Solving (vi)

$5 (x - y + z) + (4x + 3y)$

Open brackets

$5 (x - y + z) + (4x + 3y) =5x - 5y +5z+4x +3y$

Collect like terms

$5 (x - y + z) + (4x + 3y) =5x +4x +3y- 5y +5z$

$5 (x - y + z) + (4x + 3y) =9x - 2y +5z$

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aleksandrvk [35]

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4 years ago

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